Final PPT

1. Watermanagement
“Recapitulation”

2. Goal Course Watermanagement
• Knowledge of the working of the water
system, how it works, what has an influence
on what, what are you aiming at, which
processes take place
• Policy

3. Goals
• Optimal circumstances for farmers and nature, optimal
water table and gain
• Prevention of water trouble, fulfill on the rainfall event
NBW, 1:25 year shower, Stowa
• Prevention of water shortage? No guidelines exist, even
for a dry period, maintain the water level?
• Water framework directive, European directive, ecological
quality. Is about the quality of the water system itself.
• Prevention of salination?

4. Model Sobek
• Area 1 500 ha
gsl +0,8 m
Depth drainage 1 m
Summer level (weir) -0,3 m , freeboard 1,1 m
• Area 2 500 ha
gsl +0,1 m
Depth drainage 0,7 m
Summer level (pump) -0,7 m , freeboard 0,8 m

5. De Zeeuw Hellinga
Higher factor,
faster discharge of
water to the ditch

6. • Infiltration 99 mm/hour !!
• Depression 5 mm
• μ=0,060 storage in soil
• Drainage De Zeeuw Hellinga
• No seepage
• Pump 11 mm/day = 1,27 m3/s
• Precipitation 96 hour 100mm, 4 days
• Calculated steps 5 min, for 1 month

7. Used Rainfall 1:25 year 96 hours,
type long, lengthened to 1 month

8. Oversight Summer
Initial GWL =
Open water level

9. Questions
• What is the percentage surface water?
• Surface water max 11,2 ha = 1,1 %
• μ=0,060 storage soil
• How much m3 of water can be stored in a
layer of 10 cm soil, in an area of 1000 ha.
• 1000 x 10.000 x 0,1 x 0,06 = 60.000 m3
• What is the amount of precipitation in m3?
• 1000 x 10.000 x 100/1000 = 1.000.000 m3
• How much water will be discharged (max)?
• 11/1000 x 4 x 1000 x 10.000 =440.000 m3

10. More questions
• Give an estimate of the storage in the
surface water.
• Average height
(1,1 + 0,8)/2 = 0,95 m
• Assume rectangular ditch
• 11,2 x 10.000 x 0,95 = 106.400 m3

11. And more questions!
• Suppose the entire area of 1000 ha is flat, will it
submerge? (simpel approach to the problem)
• Rainfall 1.000.000 m3
• Discharge pump 440.000 m3
• Storage surface water 106.400 m3
• Storage in soil 0,95 x 10 x 60.000 = 570.000 m3
• Total of storage + discharge =1.116.400 m3
• Answer is no, differences in height are needed, so
the water can flow to the lowest points.

12. Calibration!!!!
• Which information do you need for a good
calibration?
• Precipitation
• Variances in water level
• GWL at the start of the rain shower
• Discharge of the pump
• Model that can be modified

13. Summer, maximum wl

14. Surface runoff
Gebied 1= 0

15. Groundwater discharge

16. Infiltration

17. Storage on land (mm)

18. Groundwaterlevel

19. Summer, larger culvert

20. Winter
• Area 1
gsl +0,8 m
Depth drainage 1 m
Winter level (weir) -0,5 m , dry height 1,3 m
GWL is 0,5 m below ground surface level
• Area 2
gsl +0,1 m
Depth drainage 0,7 m
Winter level (pump) -0,9 m , dry height 1,0 m
GWL is 0,5 m below ground surface level

21. Winter, maximum wl

22. 10 ha open water extra downstreams

23. Winter pump capacity 2 m3/s

24. Water level lowered 0,2m +
5000 m2 storage

25. Small culvert round 600 mm

26. Other possibilities
• Heighten the land
• Discharge partly to another area.
• Less drainage?

27. Effects of bad maintenance
• Culverts closed with dirt
• Pump capacity is lower
• Collapsed banks, lower storage and lower
discharge
• Plants growing in ditch (problem?)